I wouldn't post mere 'problems' here normally, but I will hopefully be starting to give chemistry education to some 10-11-12th graders and I want to make sure that it is not I that has a problem but rather the problem in the book itself. Here is the problem:

$100~\mathrm{mL}$ of $\ce{H2SO4}$ solution is titrated with $5\cdot 10^{-2}\ \mathrm{M}$ solution of $\ce{NaOH}$. Here are the information that are given in the titration graph:

Total volume of $\ce{NaOH}$ solution added: $0$ | pH: $x$

Total volume of $\ce{NaOH}$ solution added: $400$ | pH: $7$

Total volume of $\ce{NaOH}$ solution added: $600$ | pH: $12$

According to what has been provided above, what is the $\mathrm{pH}$ of the initial $\ce{H2SO4}$ solution ($x$) ? (Assume that $\ce{H2SO4}$ fully dissociates in water to give $\ce{2H+}$ and $\ce{SO4^2-}$ ions)

A) 1

B) 2

C) 3

D) 4

E) 5

My answer: $-\log (2\cdot 10^{-1})$

My steps:

  1. The $\ce{H+}$ ions in the acid solution have been neutralized by $400\ \mathrm{mL}$ of $5\cdot 10^{-2}\ \mathrm{M}\ \ce{NaOH}$ solution. That means that the amount of $\ce{OH-}$ ions in $400~\mathrm{mL}$ of that $\ce{NaOH}$ solution must equal the amount of $\ce{H+}$ ions in the acid solution.

  2. The amount of $\ce{OH-}$ ions in $400~\mathrm{mL}$ of $5\cdot 10^{-2}~\mathrm{M}\ \ce{NaOH}$ solution can be found by taking the product of the liter of solution and the molarity of the solution. And that gives us $0.4\times 5\cdot 10^{-2} = 2\cdot 10^{-2}\ \mathrm{mol}\ \ce{OH-}$ ions. (Yes, I didn't use the units, I don't have to, I explained with words what I was going to do)

  3. Since we used $2\cdot 10^{-2}\ \mathrm{mol}\ \ce{OH-}$ ions to neutralize the acid solution, that should mean that there were $2\cdot 10^{-2}\ \mathrm{mol}\ \ce{H+}$ in the acid solution.

  4. To find the pH of the acid solution, we need to find the concentration of $\ce{H+}$ ions in that solution. The volume of the solution is known, $100~\mathrm{mL}$. And the amount of $ce{H+}$ ions too is known, $2\cdot 10^{-2}\ \mathrm{mol}$. The concentration of $\ce{H+}$ ions in the acid solution should therefore be $2\cdot 10^{-2} / 0.1 = 2\cdot 10^{-1}\ \mathrm{M}$.

  5. $-\log[\ce{H+}] = -\log (2\cdot 10^{-1}) =$ none of the choices above.


Your calculations are correct. I also did a second sanity check: Adding $200~\mathrm{ml}\ \ce{NaOH}$ to a neutral solution does indeed give a $\mathrm{pH}\ 12$ solution, so they also wanted you to assume that the neutral point is the equivalence point.

Which answer to select? Well, considering that:

$$-\log(2 \cdot 10^{-2}) \approx 1.699 \approx 2$$

I would probably go with $2$. Along with telling the pupils that this is a badly worded question and in their exam there will be clearly solveable ones.

  • $\begingroup$ This could also be viewed as an exercise in significant figures. There is just one significant figure in $5\cdot 10^{-2}\ \mathrm{M}$ solution of $\ce{NaOH}$ and one significant figure in a pH of 7. $\endgroup$ – MaxW Aug 21 '16 at 15:37
  • $\begingroup$ @MaxW True, but there is only one sig fig in $2\log 2$, too. $\endgroup$ – Jan Aug 21 '16 at 15:41
  • $\begingroup$ that was my point. For one significant figure 1.699 rounds to 2. So it doesn't seem that the question is really "badly worded." $\endgroup$ – MaxW Aug 21 '16 at 15:50
  • $\begingroup$ @MaxW Hmm … When in doubt, I would prefer the single sig fig $2\log 2$ over the other single sig fig … Fell free to add your view of the topic as an additional answer, though ;) $\endgroup$ – Jan Aug 21 '16 at 15:52

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